3.648 \(\int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {3+2 \cos (c+d x)}} \, dx\)

Optimal. Leaf size=58 \[ \frac {2 \sqrt {-\tan ^2(c+d x)} \cot (c+d x) F\left (\left .\sin ^{-1}\left (\frac {\sqrt {2 \cos (c+d x)+3}}{\sqrt {5} \sqrt {\cos (c+d x)}}\right )\right |-5\right )}{d} \]

[Out]

2*cot(d*x+c)*EllipticF(1/5*(3+2*cos(d*x+c))^(1/2)*5^(1/2)/cos(d*x+c)^(1/2),I*5^(1/2))*(-tan(d*x+c)^2)^(1/2)/d

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Rubi [A]  time = 0.05, antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.040, Rules used = {2815} \[ \frac {2 \sqrt {-\tan ^2(c+d x)} \cot (c+d x) F\left (\left .\sin ^{-1}\left (\frac {\sqrt {2 \cos (c+d x)+3}}{\sqrt {5} \sqrt {\cos (c+d x)}}\right )\right |-5\right )}{d} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[Cos[c + d*x]]*Sqrt[3 + 2*Cos[c + d*x]]),x]

[Out]

(2*Cot[c + d*x]*EllipticF[ArcSin[Sqrt[3 + 2*Cos[c + d*x]]/(Sqrt[5]*Sqrt[Cos[c + d*x]])], -5]*Sqrt[-Tan[c + d*x
]^2])/d

Rule 2815

Int[1/(Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[(-2*
Sqrt[a^2]*Sqrt[-Cot[e + f*x]^2]*Rt[(a + b)/d, 2]*EllipticF[ArcSin[Sqrt[a + b*Sin[e + f*x]]/(Sqrt[d*Sin[e + f*x
]]*Rt[(a + b)/d, 2])], -((a + b)/(a - b))])/(a*f*Sqrt[a^2 - b^2]*Cot[e + f*x]), x] /; FreeQ[{a, b, d, e, f}, x
] && GtQ[a^2 - b^2, 0] && PosQ[(a + b)/d] && GtQ[a^2, 0]

Rubi steps

\begin {align*} \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {3+2 \cos (c+d x)}} \, dx &=\frac {2 \cot (c+d x) F\left (\left .\sin ^{-1}\left (\frac {\sqrt {3+2 \cos (c+d x)}}{\sqrt {5} \sqrt {\cos (c+d x)}}\right )\right |-5\right ) \sqrt {-\tan ^2(c+d x)}}{d}\\ \end {align*}

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Mathematica [B]  time = 1.05, size = 140, normalized size = 2.41 \[ \frac {4 \sqrt {\cos (c+d x)} \sqrt {2 \cos (c+d x)+3} \sqrt {-\cot ^2\left (\frac {1}{2} (c+d x)\right )} \csc (c+d x) F\left (\left .\sin ^{-1}\left (\frac {\sqrt {(2 \cos (c+d x)+3) \csc ^2\left (\frac {1}{2} (c+d x)\right )}}{\sqrt {6}}\right )\right |6\right )}{d \sqrt {-\cos (c+d x) \csc ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {(2 \cos (c+d x)+3) \csc ^2\left (\frac {1}{2} (c+d x)\right )}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[Cos[c + d*x]]*Sqrt[3 + 2*Cos[c + d*x]]),x]

[Out]

(4*Sqrt[Cos[c + d*x]]*Sqrt[3 + 2*Cos[c + d*x]]*Sqrt[-Cot[(c + d*x)/2]^2]*Csc[c + d*x]*EllipticF[ArcSin[Sqrt[(3
+ 2*Cos[c + d*x])*Csc[(c + d*x)/2]^2]/Sqrt[6]], 6])/(d*Sqrt[-(Cos[c + d*x]*Csc[(c + d*x)/2]^2)]*Sqrt[(3 + 2*C
os[c + d*x])*Csc[(c + d*x)/2]^2])

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fricas [F]  time = 1.61, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {2 \, \cos \left (d x + c\right ) + 3} \sqrt {\cos \left (d x + c\right )}}{2 \, \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cos(d*x+c)^(1/2)/(3+2*cos(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(2*cos(d*x + c) + 3)*sqrt(cos(d*x + c))/(2*cos(d*x + c)^2 + 3*cos(d*x + c)), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {2 \, \cos \left (d x + c\right ) + 3} \sqrt {\cos \left (d x + c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cos(d*x+c)^(1/2)/(3+2*cos(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate(1/(sqrt(2*cos(d*x + c) + 3)*sqrt(cos(d*x + c))), x)

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maple [B]  time = 0.22, size = 116, normalized size = 2.00 \[ -\frac {\sqrt {2}\, \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \sqrt {10}\, \sqrt {\frac {3+2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}, \frac {i \sqrt {5}}{5}\right ) \left (\sin ^{4}\left (d x +c \right )\right )}{5 d \sqrt {3+2 \cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{\frac {3}{2}} \left (-1+\cos \left (d x +c \right )\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/cos(d*x+c)^(1/2)/(3+2*cos(d*x+c))^(1/2),x)

[Out]

-1/5/d*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)/(3+2*cos(d*x+c))^(1/2)*10^(1/2)*((3+2*cos(d*x+c))/(1+cos(d*x+
c)))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),1/5*I*5^(1/2))*sin(d*x+c)^4/cos(d*x+c)^(3/2)/(-1+cos(d*x+c))^2

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {2 \, \cos \left (d x + c\right ) + 3} \sqrt {\cos \left (d x + c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cos(d*x+c)^(1/2)/(3+2*cos(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

integrate(1/(sqrt(2*cos(d*x + c) + 3)*sqrt(cos(d*x + c))), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {1}{\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {2\,\cos \left (c+d\,x\right )+3}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^(1/2)*(2*cos(c + d*x) + 3)^(1/2)),x)

[Out]

int(1/(cos(c + d*x)^(1/2)*(2*cos(c + d*x) + 3)^(1/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {2 \cos {\left (c + d x \right )} + 3} \sqrt {\cos {\left (c + d x \right )}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cos(d*x+c)**(1/2)/(3+2*cos(d*x+c))**(1/2),x)

[Out]

Integral(1/(sqrt(2*cos(c + d*x) + 3)*sqrt(cos(c + d*x))), x)

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